Determine if $G$ is a group under the $\,\gcd\,$ operation 
Let $G = [1,2,3,4,6,12].\;$ Let $\,a*b = \gcd(a,b), a,b \in G.\;$ Determine whether $G$ is a group.

I have found that for any two elements in $G$, commutativity holds, but the inverses are not unique for some elements i.e. $\gcd(3,4) = \gcd(3,2) = 1$, and the identity is not unique, e.g., $\gcd(2,2) = 2$ but $\gcd(3,3) = 3$. 
So $G$ is clearly not a group, is my logic correct?  
 A: Yes...the failure of any one of the group axioms is sufficient reason to conclude $G$ is not a group.
But I think it's worth the time and effort you put in to explore all the ways in which it fails to be a group. 
Note: the only element that could possibly be the identity such that $ge = eg = g$ for all $g \in G$ is the element $\;e=12:\; \gcd(g, 12) = \gcd(12, g) = g$ for all $g \in G$, and then knowing that, you can show that for all $\,g\in G g\neq 12$, there does not exist any $g^{-1} \in G$.
A: Further to the comments above, you want to check each of the group axiomse


*

*It is fairly straightforward to check that associativity holds.

*Contrary to those comments above, you should be able to find an element that is the identity. 

*Once you have worked out what this identity element is, then (and only then) are you in a position to prove that there are no inverse elements).

A: What @amWhy pointed is enough and I don't know if you are online to see my post or not. Other points that I can add: 
Your structure is closed under the given operation and indeed because of the nature of operation gcd, we have a*b=b*a:
gcd(1,1)=1   gcd(1,2)=1   gcd(1,3)=1   gcd(1,4)=1   gcd(1,6)=1   gcd(1,12)=1   

gcd(2,2)=2   gcd(2,3)=1   gcd(2,4)=2   gcd(2,6)=2   gcd(2,12)=2

gcd(3,3)=3   gcd(3,4)=1   gcd(3,6)=3   gcd(3,12)=3

gcd(4,4)=4   gcd(4,6)=2   gcd(4,12)=4

gcd(6,6)=6   gcd(6,12)=6

gcd(12,12)=12

And we can easily check that $(G,*)$ is an associative structure so it is a finite semigroup under this operation. You have more about $G$, that is $G$ is a regular semigroup. Indeed, for every element $a\in G$ the set of elements $x\in G$ in which $a*x*a=a$of is not empty and in fact we have: $$a\in\{x\in G\mid \gcd\left(\gcd(a,x),a\right)=a\}.$$
